PHP - Insert

I am wondering about the following problem:
I have two sessions, one for the user ($_SESSION['user_id']) and second one for the products inside cart ($_SESSION['cart']). I need to insert data into table racun.
I did it this way and it works, but i want to know if this can be done better?
Code: [Select]
while ($row=mysql_fetch_array($query)){
$id = $_SESSION['korisnik_id'];
$idp =  $row['product_id'];
$kol = $_SESSION['cart'][$row['product_id']]['quantity'] ."<br>";
$insert = "INSERT INTO racun (product_id, quantity, korisnik_id) VALUES ('$idp', '$kol', '$id') ";
$result = mysql_query($insert);
  }

Hey, who can help?

Example:
I want to insert a row with a input and a select option. I am inserting now with select option, but i want to insert with a input too, because if i need to insert a link instead of choosing from select option.
---------
(puts a link)Link: ...link...
(chooses nothing)Select: Choose one...
Submit
---------
I need something like if LINK is filled insert it. Nothing happens to SELECT, but they have to have the same row names.
Hope its clear enough.

Cheers

Would someone be able to tell me why I can't run a successful
mysql_query($sql)
with
$sql="INSERT INTO myDB.Titles (Title, Year, cID) values('". $title . "', '" . $year . "', '" . $cID . "')";

$sql echoes out to "INSERT INTO myDB.Titles (Title, Year, cID) values('Test', '2000', '2')"
It looks right but I'm probably off on the quotes somewhere. Thanks in advance.

Just a quick question I have a form, couple of drop down boxes, input fields etc.. at the end is the submit button. now my question is should my insert into database sqls go before the submit button inside the form?
out side the form? after the button inside the form?

If by any chance you could give me some advice on the best way to layout my sql that would be great. They are big and go into three tables, at the moment I have them as three seperate queries but I think they must be able to go into one. can someone please help me with the layout and syntax.

here is what the queries look like at the moment:


if ($IBselect=$_POST ['IBselect']);
{
//CUSTOMER
$enterCust="INSERT INTO customer(username, password, title, firstName, lastName, address, town, country, postCode, phone, email, dateCust) 
                        VALUES ('$_POST[username]', '$_POST[password]', '$_POST[title]', '$_POST[firstName]', '$_POST[lastName]', '$_POST[address]', '$_POST[town]', '$_POST[country]', '$_POST[postCode]', '$_POST[phone]', '$_POST[email]', 'DATE: Auto CURDATE()', CURDATE()";
$enterCust_query=mysql_query($enterCust)or die(mysql_error());

//CARD 
$enterCard="INSERT INTO card(cardNumber,  name, expDate, cardID) 
                        VALUES ('$_POST[cardNumber]','$_POST[name]', ' $_POST[expDate]', '$_POST[cardID]')";
$enterCard_query=mysql_query($enterCard);

//BOOKING
$RCenterBook=mysql_query("INSERT INTO booking (custNumber, cardID, rideName, seatNo1, seatNo2, price, price2, dateBook, ID_time_tbl) 
                VALUES ('$custNumber', '$_POST[cardID]', '$rideName', '$_SESSION[IBextract]', '$_SESSION[IBextract2]', '$IBendPrice1', '$IBendPrice2', 'DATE: Auto CURDATE()', CURDATE()',  '$_SESSION[IB_slot_Time]'");
$IBenterBooking_query=mysql_query($IBenterBook); 
        if (!mysql_query($enterCard, $enterCust, $IBenterBook))
        {
        die("Error:" .mysql_error());
        }
        echo "1 record added IB";
}


Thanks =)

Error:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'like (comment_id, name, like) VALUES ('102','Steven','1')' at line 1

$sql="INSERT INTO like (comment_id, name, like)
VALUES
('$_POST[comment_id]','$_POST[name]','$_POST[like]')";
  
if (!mysql_query($sql,$con))
  {
  die('Error:' . mysql_error());
  }
echo "You have succesfully liked this comment.<br /><INPUT TYPE=\"button\" VALUE=\"Back\" onClick=\"history.go(-1);\">";

mysql_close($con)

Table:
comment_id name like
For some reason, I cannot spot this error. Can you find it? My $con = mysql_connect is a part I left out but it connects fine! The data form is echo'ing fine but it's not inserting into the MySQL Table.

Thanks in Advanced!

Not being able to upload images in the directory with a path name in the database?

1) This is my code for insert into the database and directory:

<?php


$host="localhost";
$username="root";
$pass="";
$db="registration";

$conn=mysqli_connect($host,$username,$pass,$db);
if(!$conn){
	die("Database connection error");
}

// insert query for register page
if(isset($_POST['ronel'])){
     $images = $_FILES['file']['name'];
  $target_dir = "uploads/";
  $target_file = $target_dir . basename($_FILES["file"]["name"]);

    // Select file type
  $imageFileType = strtolower(pathinfo($target_file,PATHINFO_EXTENSION));

  // Valid file extensions
  $extensions_arr = array("jpg","jpeg","png","gif","pdf");

  // Check extension
  if( in_array($imageFileType,$extensions_arr) )
  {

$details=$_POST['details'];
$location=$_POST['location'];
$checkbox=$_POST['checkbox'];
$injured=$_POST['injured'];    
$agegender=$_POST['agegender'];   
$contact=$_POST['contact'];
$empid=$_POST['empid'];
$dept=$_POST['dept'];    
$organization=$_POST['organization'];     
$summary=$_POST['summary']; 
$name=$_POST['name'];
$outcome=$_POST['outcome'];    
$cause=$_POST['cause'];     
$action=$_POST['action'];
$reportedname=$_POST['reportedname'];
$position=$_POST['position'];    
$organisation=$_POST['organisation'];
$reportedcontact=$_POST['reportedcontact'];
$reporteddept=$_POST['reporteddept'];    
$status="Pending";
$comment=$_POST['comment'];
	
	
	 $query="INSERT INTO `proposals` (`details`,`location`,`date`,`time`,`checkbox`,`injured`,`agegender`,`contact`,`empid`,`dept`  ,`organization`,`summary`,`image`,`outcome`,`cause`,`action`,`reportedname`,`position`,`organisation`,`reportedcontact`,`reporteddept`,`status`,`comment`) VALUES ('$details','$location', current_timestamp(),current_timestamp(),'$checkbox','$injured','$agegender','$contact','$empid','$dept' ,'$organization','$summary','$name','$outcome','$cause','$action','$reportedname','$position','$organisation','$reportedcontact','$reporteddept','$status','$comment')";
	
	$res=mysqli_query($conn,$query);
	if($res){
		$_SESSION['success']="Not Inserted successfully!";
		header('Location:');
	}else{
		echo "<script>alert('Proposal not applied!');</script>";
	}
      // Upload file
     move_uploaded_file($_FILES['file']['tmp_name'],$target_dir.$image);
	}
}
date_default_timezone_set("Asia/Kolkata");
?>

2) Here is the input file:

<form class="form-horizontal" method="post" action="" enctype="multipart/form-data">
            <input type="hidden" name="ronel" value="">
  <div class="form-group">
  <label style="position:absolute; left:63%; top:425px;" for="inputEmail" class="col-lg-3"><b>Upload Images Here :</b></label><br><br>
  <div class="col-lg-9">
  <input style="position:absolute; left:78%; top:420px;" type="file" name="file" enctype="multipart/form-data" class="form-control" name="incident_reference" onchange="document.getElementById('inc_ref').src = window.URL.createObjectURL(this.files[0]); document.getElementById('inc_ref').className +='_active'; document.getElementById('inc_ref_span').className += '_hidden'">
  </div><iframe id="inc_ref" class="form-group" width="220px" height="130px" style="position:absolute; left:78%; top:32%;"></iframe></div>
</form>

3) error code:

Notice: Undefined index: name in /opt/lampp/htdocs/create-nearmiss.php on line 57
ONGC TRIPUR

 

Edited March 28, 2020 by Ronel
change of var

Dear Sir/Madame

I am trying to insert comments into a table called comments using echo ID in the input form but so far i can insert the comments with the post id but unable re-comment again kindly help me what do i do? 

This is the input form:

 <div>
        <form action="ehscomment.php" method="post">
        <input type="hidden" name="id" value="<?php echo $id ?>"> 
            <div>
                <label>Add comment</label>
    <div>
        <textarea rows="6" cols="110" name="comment" placeholder="comment"></textarea>
    
        </div>
        </div>
            <input type="submit" name="postcomment" value="comment"></form>
</div> 

And this is ehscomment.php

<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "registration";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
   die("Connection failed: " . $conn->connect_error);
} 
if(isset($_POST['postcomment'])){
       $id = $_POST['id'];
       $comment = $_POST['comment'];
       

$sql = "INSERT INTO comments (id,comment)
VALUES ('$id','$comment')";

if ($conn->query($sql) === TRUE) {
   echo "New record created successfully";
} else {
   echo "Error: " . $sql . "<br>" . $conn->error;
}

$conn->close();}
?>

Error message: Error: INSERT INTO comments (id,comment) VALUES ('242','asddd')
Duplicate entry '242' for key 'PRIMARY'

i am trying to update a column in the database. I can update int columns but what about text? i want to update it so that the new text is added to it while keeping the old one. Here is what i have(which isnt working):

mysql_query("UPDATE ".DB_PREFIX."posts
            SET post_content = '$old_content' + '".$_POST['test']."'
            WHERE post_id = '$double_post_id'") or die(mysql_error());

i have also tried:

SET post_content = post_content + '".$_POST['test']."' 

but that didnt work either. What should i be doing?

Okay,
I'm hoping one of you can help me.
I have a mysql database that I have configured through phpmyadmin.
I have an android app that simply makes and sends a mysql query
I can get it to successfully return values when using Select statements but when I use INSERT INTO, it returns " Error Query is invalid" BUT BUT BUT, when I use the same string and enter it through the sql tab in myphpadmin it works fine !

So here is the string ( the semicolons at the end of each field name are so I can use something common to split the string up when the data arrives back on the phone)
Code: [Select]
randomkey||||||INSERT INTO table4  (`geolat;` , `geolong;` , `mode;` , `destgeolat;` , `destgeolong;` , `cellphone;` , `email;` , `carrego;` , `colour;` ,  `rating;` , `comment;`)
VALUES (0.0,0.0,'driver' ,-43.54779,172.62472, , '' ,'' , 'text' , 'ratingleftblank' , 'commentblank' )
the index4.php script is as follows
Code: [Select]
?php
/*
 * Written By:
 * James
 */


/************************************CONFIG****************************************/
//DATABSE DETAILS//
$DB_ADDRESS="mysql1.openhost.net.nz";
$DB_USER="bling44";
$DB_PASS="sadlyinept";
$DB_NAME="bling44";

//SETTINGS//
//This code is something you set in the APP so random people cant use it.
$SQLKEY="randomkey";
/************************************CONFIG****************************************/

//these are just in case setting headers forcing it to always expire and the content type to JSON
header('Cache-Control: no-cache, must-revalidate');
header('Content-type: application/json');


if(isset($_POST['tag'])){ //checks ifthe tag post is there
$tag=$_POST['tag'];
$data=explode("||||||",$tag); //split the SQL statement from the SQLKEY
if($data[0]==$SQLKEY){ ///validate the SQL key
$query=$data[1];
$link = mysql_connect($DB_ADDRESS,$DB_USER,$DB_PASS); //connect ot the MYSQL database
mysql_select_db($DB_NAME,$link); //connect to the right DB
if($link){
$result=mysql_query($query); //runs the posted query (NO PROTECTION FROM INJECTION HERE)
if($result){
if (strlen(stristr($query,"SELECT"))>0) { //tests if its a select statemnet
$outputdata=array();
while ($row = mysql_fetch_assoc($result)){
$outputdata[]=$row; //formats the result set to a valid array
}
echo json_encode(array("VALUE",$tag,array_merge($outputdata))); //sends out a JSON result with merged output data
} else {
echo json_encode(array("VALUE",$tag,array_merge(array(array("AFFECTED_ROWS ".mysql_affected_rows($link)))))); //if the query is anything but a SELECT it will return the array event count
}
} else echo json_encode(array("VALUE",$tag,array_merge(array(array("ERROR QUERY IS INVALID"))))); //errors if the query is bad
mysql_close($link); //close the DB
} else echo json_encode(array("VALUE",$tag,array_merge(array(array("ERROR Database Connection Failed"))))); //reports a DB connection failure

} else {
echo json_encode(array("VALUE",$tag,array_merge(array(array("ERROR BAD CODE SUPPLIED"))))); //reports if the code is bad
}
}
?>
So to reiterate.
I can search the DB but can't INSERT INTO, unless I go through the myphpadmin interface.
Any ideas are very much appreciated

hey developers,   i have problem to get last insert id. btw.. i try to put that code on my mvc creation.   on register_model page i write like this:  public function numbering(){
        $sth = $this->db->prepare('SELECT * FROM member_data');
        $result = $this->db->lastInsertId();
        return $result;
    }
  and for the controllers page, i write the code like this:   $c = $this->model->numbering();
         foreach ($c as $key => $value) {
$data = $result('id');             
}          
        
$abc['member_id'] = $data;     but i didn't get the results,.. can you help me?

I'm trying to use PDO and get used to doing things this way. I've been away from php/mysql for a few years, so, I'm crusty.  I'm not getting any error messages back on this code, but the insert just doesn't happen.  My first guess is that I'm doing something wrong with the datetime now() function.  But, I may not have the PDO code right. I tried the script the old fashion way with mysql_query() and that worked.  So, it has to be something in this code. I believe my server is set up to do PDO as it shows:
PDO
PDO support   enabled
PDO drivers    mysql, sqlite

pdo_mysql
PDO Driver for MySQL, client library version   5.0.45

My php version is 5.2.14 and Mysql is 5.0.45.

Any help would be appreciated.

Code: [Select]
     $DBH = new PDO("mysql:host=$host;dbname=$dbname", $user, $pass);
     $DBH->setAttribute( PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION );

     $sql=$DBH->prepare("INSERT INTO assets(asset_name,date_added,short_desc) VALUES (:asset_name,NOW(),:short_desc)");

     $sql->bindParam(':asset_name',$asset_name);
     $sql->bindParam(':short_desc',$short_desc);

     $name=$_POST["input1"];
     $short_desc=$_POST["input2"];

     $DBH->exec();
     echo $name;

    echo "\nPDO::errorInfo():\n";
    print_r($DBH->errorInfo());
   }
   catch(PDOException $e) {
     echo "Syntax Error: ".$e->getMessage();
   }

Hello!Is this correct?

Code: [Select]
$sql="INSERT INTO pacienti (nume, prenume, cnp, varsta, sex, casa_asigurari, oras, strada, numar, judet, data_inregistrarii)
VALUES
('$_POST[nume]','$_POST[prenume]','$_POST[CNP]','$_POST[varsta]','$_POST[sex]','$_POST[casa]','$_POST[oras]','$_POST[strada]','$_POST[nr]','$_POST[judet]','$_POST[internare]')";

$sql="INSERT INTO diagnostic (nume)
VALUES
('$_POST[diagnostic]')";
and i mean..if i can do 2 insert in 2 different tables...or how can i do this?

hi;


Code: [Select]
$giden = $_GET['sehir_part']+1;
$url = '31.php?sehir_part='.$giden.'';
//$erkek_top = 7
for($q=1;$q<=7;$q++)
{

//i want insert into to mysql in here ; but its doing only one inserting
so i want be 7 more insert

}
header( 'refresh:1;url=31.php?sehir_part='.$giden)and in every page loading i want inserting to mysql againg

pls helpme thanks

could someone help me with this.  here is what I have:

<?php require_once('dbinfo.php'); ?>

<?php

mysql_connect($dbaddress,$username,$password);
mysql_select_db($db) or die("Cannot find database!");

/*
$query = "SELECT * FROM contacts";
$result = mysql_query($query);
$numRows = mysql_numrows($result);
$i = 0;

while ($i < $numRows) {
echo mysql_result($result, $i, "first") . " " . mysql_result($result, $i, "last");
$i++;
  }

*/

$fname = 'WrdScramble()';
$ftype = 'DAO';
$fdesc = 'Randomly rearranges the characters in a string or word.';
$fcode = '';

$query = "INSERT INTO functions VALUES ('','$fname','$fcode','$ftype','$fdesc')";
mysql_query($query);
print($query);

//$query = "SELECT fcode FROM contacts WHERE first = 'code'";
//$result = mysql_query($query);
//$numRows = mysql_numrows($result);
//$fcode = mysql_result($result, 0, "fcode");

?>


<?

//close connection
mysql_close();

?>


It is not inserting the data into the database, and if I use:

mysql_error()

and print it out, I get nothing with that either.  I am not sure where to go!  I also have used this:

(fid, fname, fcode, ftype, fdesc)

preceeding the 'VALUES' word in the statement just to see if it would take.  No luck!  Do I not see the obvious?  Any help appreciated!  thanks!  (I have checked my field types and max length in PHPmyadmin and they are not an issue.)

Quote

$con = mysql_connect("localhost","****","****");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("checkmyw_database", $con);

mysql_query("INSERT INTO requests (username, password, email, weavecode, mobile, jobtitle, relationship, sexuality, venue, type)
VALUES ('$me', '$pw', '$email','1999','$mobile','$job','$relationship','$sexuality','$venue','newuser')");


mysql_query("INSERT INTO members (username, password, email, weavecode, mobile, jobtitle, relationship, sexuality,)
VALUES ('$me', '$pw', '$email','1999','$mobile','$job','$relationship','$sexuality',)");



mysql_close($con);





any idea why this doesn't work?